# Cancellation of forces in feeling weight

If we are on the ground, Earth's center attracts us by $$mg$$ force and the ground also exerts $$mg$$ force in the upward direction. Those two forces cancel out. So there is no net force acting on us now. Then how do we feel weight and not weightlessness? Since net force is $$0$$, we shouldn't feel any weight.

• Does this answer your question? If gravity isn't a force, then why do we learn in school that it is? Jun 13, 2021 at 23:19
• There's no NET force acting, but there are the two forces you mentioned. Jun 14, 2021 at 2:16
• If only the $mg$ from the earth acted you would be in free fall. It is you sensing the counter force your body creates that is weight. Jun 14, 2021 at 13:28

## 6 Answers

There is a force. Try forcing your two hands together very hard. They cancel each other but you can still feel the force or the pressure. When you apply more force they still cancel but the pressure grows. Like the Earth a force is meeting a resistance and you feel it. In free fall the force is there but there is no resistance to feel.

• Similarly, if you pull on the end of a piece of rope equally to put tension on it, it doesn't mean there is no force in the rope just because the forces cancel out. Or perhaps, more illustrative, imagine if someone put you on a torture racked and pulled your arms and legs in opposite directions. You definitely feel a force even if your legs and arms are still attached. Jun 13, 2021 at 20:02
• Another example is inertia. When you fire a rocket in space a force is felt when all atoms or subatomic parts squeeze up against each other. Again the resistance is felt. If the rocket maintains the same speed eventually the parts spread back out and we feel less and less resistance until eventually there is no force again. Jun 13, 2021 at 20:11

Since net force is 0,we shouldn't feel any weight.

This kind of statement can sound right intuitively, but actually has no basis in fact.

If two trucks hit you from opposite sides, then there's 0 net force on you. But you'll still probably feel something!

For a human to "feel" a force really means for something to press up against their nerves. Gravity applies to each molecule equally, so nothing presses against your nerves. This is why when you're in free fall, you feel nothing even though gravity is still pulling $$9.8 m/s^2$$. When you're resting in a chair, you feel the normal force of the chair because it applies to different parts of your body unevenly, and so certain nerves will be compressed to various degrees.

The sensed pressures Bill Alsept describes are called proprioceptive and the nerve impulses generated in this way tell you where your arms and legs are positioned even when you cannot see them. Proprioception also tells you where down is relative to your body; in this case, the pressures sensed by the nerves are caused by the pull of gravity on your muscles and bones.

Also in free fall when there is a force on you, you feel weightless - isn't that ironic!

The reason is that there are relative stresses that you experience in your body that are not alleviated by the normal force. While the normal reaction is digging into your heals, gravity is trying to squeeze you from your head down, into the ground. Do you see how there is relative stress in the body? All our muscles and tissue are "hanging-off" our skeleton which stays in place and supports them. All these stresses induce the feeling of weight.

Imagine a flexible massive rod with masses at the end. Balance this rod on your finger tip. Is the net force on the rod zero ? Yes. But do you see the rod droop at the edges, slightly bent under the weight of the end masses. This relative stress is what constitutes the feeling of weight.

On a deeper level, is the net force really zero when you stand ? Is there really a force when you free fall?
Happy GR!

If the sum of forces on a non rotating body is zero, that means zero acceleration, or: the body is at rest or with constant velocity regarding an inertial observer. Only in this sense $$\sum \mathbf F_i = 0$$ is equivalent to no force at all.

But there is another way to measure force, besides acceleration: elastic displacement from the Hooke's law: $$F = -kx$$. That is the way load cells work. When we are in a vertical position, the distances between several bones (and the length of some bones themselves) get shorter, (what can be painful if the cartilages of the knees are not in good state for example). This compressive strain is transformed in electric impulses to our brain. Our feeling of a compressive force results from the way our body measures that strains.

Why can’t we say the forces cancel each other ?

When you apply a force or push of 150 N on the wall. You will notice you move back. According to your statement , you should have had not moved. The point is , when you applied 150 N on wall. The wall also applied 150N on you. There are two forces. One is acting on wall and the other on you. Both have opposite direction but not the same body. Therefore , they cant cancel each other.

Like here : let’s say your mass = 50 kg and acc you applied on wall = $$3 m/s^2$$. If mass of the wall is $$10^6$$ kg. Then ,

$$50 * 3 = 10^6 * Acc_{wall} = 150 N .$$

Therefore , you notice that actually the acc of wall is very small.

Same is with earth. Acc of earth by is almost negligible evasive of such a huge mass of earth.

Edit : When is net force = 0 ?

Check what is free body diagram.

Brief definition : If you are writing forces acting on the body , then we do not write the forces which are acting by the body on other bodies.

If I keep my book on table. Forces acting on book are force by the earth downwards and normal force upwards.

NOTE : The force by book on earth is acting on earth. Not on book.

Therefore +N -mg(sing convention from Cartesian coordinates) = mass of book * acc of book(0). Therefore , N and mg can cancel out or N=mg.

I hope it is clear.

• Are you saying that while standing at rest on earth, net force isn't zero, in the non-gr picture? If so, thats wrong. Jun 14, 2021 at 20:54
• Same is with earth. No. In your wall pushing example, there is one and only one force - the wall's normal reaction. But in the case of standing at rest, there is gravity in addition and they both count since they are external. Jun 14, 2021 at 20:57
• @lineage No. Read the answer carefully. When did I write net force isn’t zero ? . Did you N- mg = 0 ? Jun 15, 2021 at 5:14
• @lineage What I wrote about wall example is that when you push the wall , you get pushed back . In this case , how can you say net force = 0 when you had an acceleration of moving back ? .Scene is different if you don’t get pushed back.In that case , the force gets absorbed in the muscles of your hand. Jun 15, 2021 at 5:15
• More over , the push you do is for a seconds. Try it on wall yourself. Jun 15, 2021 at 5:16